Connect lines per attached sketchWe have GM=GB=GN => MBN is a right triangleIn right triangles AEC and O1GO2 we have GB^2=BO1.BO2 and BE^2=BA.BCSo BG= ½ BE and MBNE is a rectangleWe have ∠ (BNE)= ∠ (BNC)=90 =>E, N, C are collinearSimilarly A,M,E are collinearNote that ∠ (BCE)= Arc(AF)+Arc(FE)And ∠ (EMN)= Arc(AF)+Arc(EH)But ∠ (EMN)= ∠ (BNM)= ∠ (BCE) => Arc( FE)=Arc(EH)=> DE is an angle bisector of angle FDH

By theorem from previous problem, AMNC is concyclic. Therefore, AM and NC must meet on radical axis DE of O1 and O2 at point P. <APC must be 90 because <MAB+<NCB=90, so P is also on big circle. P must coincide with E as result.<ACN=<AMF=<ACF+<EDH, but also<ACN=(<ADF=<ACF)+<FDE.Therefore <ACF+<EDH=<ACF+<FDE, so <EDH=<FDE